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3 years ago

How often is water added to the Earth system?

1 answer:

4 0

Water is never added to earth system. Water forever remains in the water cycle on earth, so it goes from the ground, to the air, to the rain, to the sea, and round and round continuously. This cycle means that there does not need to be new water added to the earth, because it recycles any water that already exists of its own accord.

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A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figur

pogonyaev

Answer with Explanation:

We are given that mass of block=0.0600 kg

Initial speed of block=0.63 m/s

Distance of block from the hole when the block is revolved=0.47 m

Final speed=3.29 m/s

Distance of block from the hole when the block is revolved= $9\times 10^{-2}m$

a.We have to find the tension in the cord in the original situation when the block has speed = $v_0=0.63 m/s$

$T=\frac{mv^2}{r}$

Because tension is equal to centripetal force

Substitute the values

$T=\frac{0.06\times (0.63)^2}{0.47}=0.05 N$

b. $v=3.29 m/s$

$T=\frac{mv^2}{r}=\frac{0.06\times (3.29)^2}{0.09}=7.2 N$

c.Work don=Final K.E-Initial K.E

$W=\frac{1}{2}m(v^2-v^2_0)$

$W=\frac{1}{2}(0.06)((3.29)^2-(0.63)^2)$

$W=0.31 J$

4 0

2 years ago

Read 2 more answers

What does a Lewis structure diagram represent?

densk [106]

Answer:

correct me if i'm wrong nut i thinks its a The atomic symbol of an element surrounded by valence electrons

Explanation:

5 0

2 years ago

Read 2 more answers

The main energy source during exercise of low to moderate intensity is

zhannawk [14.2K]

Answer:

THUS,THE MAJOR SOURCES OF ENERGY DURING EXERCISE ARE CARBOHYDRATES AND FATS.

Explanation:

6 0

2 years ago

HELP... 3.00 amu = _____ Mev. 3.22 x 10-3 2.79 x 103 3.10 x 102

DanielleElmas [232]

The correct answer is:
$2.79 \cdot 10^3 MeV$
Let's see why.

1 amu corresponds to the mass of the proton, which is:
$m_p = 1.66 \cdot 10^{-27} kg$
if we convert this into energy, using Einstein equivalence between mass and energy, we find:
$E=mc^2 = (1.66 \cdot 10^{-27} kg)(3\cdot 10^8 m/s)^2 = 1.49 \cdot 10^{-10} J$
Now we can convert it into electronvolts:
$E= \frac{1.49 \cdot 10^{-10}kg}{1.6 \cdot 10^{-19} J/eV} =9.34 \cdot 10^9 eV = 934 MeV$

So, 1 amu = 934 MeV. Therefore, 3 amu corresponds to 3 times this value:
$3 amu = 3 \cdot 934 MeV \sim 2790 MeV = 2.79 \cdot 10^3 MeV$

5 0

3 years ago

Read 2 more answers

A chair of mass 12.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force F

yarga [219]

Answer:

Explanation:

mass, m = 12 kg

Force, F = 40 N

θ = 37° below the horizontal

(a)

Diagram is attached

(b) Let N be the normal reaction

According to the diagram

N + F Sin θ = m g

N = mg - F Sin θ

N = 12 x 9.8 - 40 x Sin 37

N = 117.6 - 24.07

N = 93.53

6 0

2 years ago